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Question Number 62761 by lalitchand last updated on 25/Jun/19
if cosθ=((cosA.cosB)/(1−cosA.cosB))   prove that Tan(θ/2)=Tan(A/2).Cot(B/2)
$$\mathrm{if}\:\mathrm{cos}\theta=\frac{\mathrm{cosA}.\mathrm{cosB}}{\mathrm{1}−\mathrm{cosA}.\mathrm{cosB}}\:\:\:\mathrm{prove}\:\mathrm{that}\:\mathrm{Tan}\frac{\theta}{\mathrm{2}}=\mathrm{Tan}\frac{\mathrm{A}}{\mathrm{2}}.\mathrm{Cot}\frac{\mathrm{B}}{\mathrm{2}} \\ $$
Commented by Prithwish sen last updated on 25/Jun/19
I think it is  cosθ = ((cosA−cosB)/(1−cosAcosB))  ((1−cosθ)/(1+cosθ)) = ((1−cosA+cosB−cosAcosB)/(1+cosA−cosB−cosAcosB))  tan(θ/(2 ))  = (((1−cosA)(1+cosB))/((1+cosA)(1−cosB)))                 =tan(A/2).cot(B/2)  waiting for your comment.
$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{cos}\theta\:=\:\frac{\mathrm{cosA}−\mathrm{cosB}}{\mathrm{1}−\mathrm{cosAcosB}} \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\theta}{\mathrm{1}+\mathrm{cos}\theta}\:=\:\frac{\mathrm{1}−\mathrm{cosA}+\mathrm{cosB}−\mathrm{cosAcosB}}{\mathrm{1}+\mathrm{cosA}−\mathrm{cosB}−\mathrm{cosAcosB}} \\ $$$$\mathrm{tan}\frac{\theta}{\mathrm{2}\:}\:\:=\:\frac{\left(\mathrm{1}−\mathrm{cosA}\right)\left(\mathrm{1}+\mathrm{cosB}\right)}{\left(\mathrm{1}+\mathrm{cosA}\right)\left(\mathrm{1}−\mathrm{cosB}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}.\mathrm{cot}\frac{\mathrm{B}}{\mathrm{2}} \\ $$$$\mathrm{waiting}\:\mathrm{for}\:\mathrm{your}\:\mathrm{comment}. \\ $$
Commented by lalitchand last updated on 25/Jun/19
is that question is wrong?? and u have applied  componendo and dividedendo after reciporcal
$$\mathrm{is}\:\mathrm{that}\:\mathrm{question}\:\mathrm{is}\:\mathrm{wrong}??\:\mathrm{and}\:\mathrm{u}\:\mathrm{have}\:\mathrm{applied} \\ $$$$\mathrm{componendo}\:\mathrm{and}\:\mathrm{dividedendo}\:\mathrm{after}\:\mathrm{reciporcal} \\ $$
Commented by Prithwish sen last updated on 25/Jun/19
I think so.
$$\mathrm{I}\:\mathrm{think}\:\mathrm{so}. \\ $$

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